Problem: Let $S$ be the set of nonzero real numbers.  Let $f : S \to \mathbb{R}$ be a function such that

(i) $f(1) = 1,$
(ii) $f \left( \frac{1}{x + y} \right) = f \left( \frac{1}{x} \right) + f \left( \frac{1}{y} \right)$ for all $x,$ $y \in S$ such that $x + y \in S,$ and
(iii) $(x + y) f(x + y) = xyf(x)f(y)$ for all $x,$ $y \in S$ such that $x + y \in S.$

Find the number of possible functions $f(x).$
Explanation: Setting $x = y = \frac{z}{2}$ in (ii), we get
\[f \left( \frac{1}{z} \right) = 2f \left( \frac{2}{z} \right) \quad (1)\]for all $z \neq 0.$

Setting $x = y = \frac{1}{z}$ in (iii), we get
\[\frac{2}{z} f \left( \frac{2}{z} \right) = \frac{1}{z^2} f \left( \frac{1}{z} \right)^2\]for all $z \neq 0.$  Hence,
\[2f \left( \frac{2}{z} \right) = \frac{1}{z} f \left( \frac{1}{z} \right)^2. \quad (2)\]From (1) and (2),
\[f \left( \frac{1}{z} \right) = \frac{1}{z} f \left( \frac{1}{z} \right)^2,\]so
\[f(x) = xf(x)^2 \quad (3)\]for all $x \neq 0.$

Suppose $f(a) = 0$ for some $a \neq 0.$  Since $f(1) = 1,$ $a \neq 1.$  Setting $x = a$ and $y = 1 - a$ in (iii), we get
\[f(1) = a(1 - a) f(a) f(1 - a) = 0,\]contradiction.  Therefore, $f(x) \neq 0$ for all $x,$ so from (3),
\[f(x) = \frac{1}{x}.\]We can check that this function works, so there is only $\boxed{1}$ solution.